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Is there a tank priority sequence for using the fuel tanks?
Is there a tank priority sequence for using the fuel tanks, or does the pilot decide?

Welcome to Av.SE!

Your question is a really good one (I'm surprised it has not been asked before). You can narrow your question to some few aircraft model as it is aircraft specific (e.g some aircraft transfer fuel from one tank to another to adjust CG in flight, some such as fighters have less tanks and possess accumulators,...)

@ManuH Thank you for your comment. I was wondering this topic. But I don't have more information aircraft models and technical properties so I couldn't add aircraft models.

Which fuel tanks are used in order of priority in aircraft?

What you're talking about, in large aircraft, is often referred to as the fuel burn schedule. Light aircraft generally do not have a fuel burn schedule though they may have have minimal requirements.
The fuel burn schedule is dictated by the aircraft design and is thus different for different aircraft designs.

Is there a sequence of priority?

Yes, and if you don't observe the fuel burn schedule you risk exceeding the structural limitations of the aircraft, and in particular wing bending moments.

Or the pilot decides?

No, the pilot does not decide usually. While there may be some leeway in the fuel burn schedules of large aircraft, the pilot or a flight engineer (old aircraft) or an automated system does not decide but rather implements a preset protocol.
As an example, here's a general outline of the fuel burn schedule for 747-100/200 aircraft. This is from memory so might not be entirely correct but is sufficiently so for example purposes:

On start, taxi, and takeoff burn tank to engine. In other words, route the fuel from main tank 1 to engine 1, main tank 2 to engine 2, etc.
Once you're established in your cruise climb, all engines burn out of the center tank if there is fuel in the center tank until the center tank fuel is exhausted (but don't burn any fuel in the center tank that is there for ballast).
Burn engines 1 & 2 from main tank 2 and engines 3 & 4 from main tank 3 until the amount of fuel in main tank 2 matches that in main tank 1 and the amount of fuel in main tank 3 matches that in main tank 4. Main tanks 2 & 3 are larger than 1 & 4. When parity is reached, burn tank to engine.
At some point transfer the fuel in reserve tank 1 to main tank 1 and reserve tank 4 to main tank 4. Likewise, if you have them, from reserve tank 2 to main tank 2 and reserve tank 3 to main tank 3. Most 747-100/200 aircraft didn't have reserve tanks 2 & 3. You have to get the fuel out of the reserve tanks before landing.

Be aware that just as there is a fuel burn schedule, there may also be a fuel loading schedule. In the 747-100/200 that schedule dictated that no fuel goes into the center tank until all other tanks are full. As I remember, the largest center tank available held 114999 lbs of fuel at 6.7 lbs/gal fuel density. Center tank size varied as ordered by the airline and changed through the production run of the aircraft.

Fortunately, it doesn't need to be committed to memory because it will be on the checklist.

"No, the pilot does not decide usually." I assume though there are overrides possible in case of an emergency?

@Mast I can't speak to automated systems or other aircraft, but in the case of 747-100/200 aircraft the system was very flexible (and was considered complicated) with the exception that once you move fuel out of a reserve tank, you can't put it back as I remember. It's not a matter of overriding but of the flight engineer throwing switches in an order other than would normally be done.

That makes perfect sense, thanks for adding that.

And some craft, like the Concorde, had the additional complexity of needing to actively adjust the aircraft CG, by pumping fuel betwen the tanks, to maintain trim in different phases of flight.  Here you had to not only worry about the fuel schedule, but you also had to be periodically moving the remaining fuel around to keep the correct attitude during flight.

My G35 Bonanza (and many other, older aircraft) used a pressure carburetor that would return excess fuel to the left main fuel cell.
This meant that no matter what fuel cell the engine was consuming fuel from, the carburetor would always be sending the fuel it did not feed to the intake manifold back to that particular fuel cell.
If the left main fuel cell were full of fuel, the returned fuel from the carburetor would be vented overboard.
The workaround for this behavior of the fuel system was to consume fuel from the left main fuel cell for at least the first hour of the flight before consuming fuel from the other fuel cells/tanks.  This would leave space in the left main fuel cell for the the carburetor return unused fuel without it being vented overboard.

Yes there is often a prescribed sequence.  Sometimes it has to do with wing bending moments; often it has to do with the fore-and-aft location of the CG.  For example my understanding is that the P-51D (and C?) Mustangs were rather unpleasant to fly while the aftmost tank was full of fuel, so some fuel from that tank would typically be consumed even before using any fuel from the drop tanks.
And sometimes it just has to do with the plumbing of the fuel system.  For example if only tank A is directly connected to the engine, you wouldn't want to forget to transfer some fuel from tank B to tank A before tank A runs dry.  Sometimes one tank feeds more reliably than others due to where the boost pump is located so it is wise (and prescribed in the manual) to use that tank for take-off.  Etc.

As others have already alluded, this really depends on the aircraft. Your question does not specifically state fixed wing - so here is my $.02
I know from flying Robinson model helicopters, specifically the R22 and R44, the main and auxiliary fuel tanks are actually tied together - you burn from both at all times; there is only a master shutoff, there is no tank selector.

It strikes me as quite odd that the tanks are different sizes. Wouldn't you empty the left tank (from the perspective of this drawing) before emptying the right tank and thus be left with an imbalance?

Yep! And that is by design! When you solo, you fly from the right seat. The imbalance offsets a solo pilot (and some other forces related to anti-torque).

Got it. I thought about that, but thought a solo sat on the left (my IANAP is showing...). What do you do when you've got a passenger (he asks as a whole new question in  the comments)?

Yes! When you are filling up the aircraft and you use the centre tanks, it then means that the pilot has to use the centre tanks first before the 1 or 2 wing tanks are selected, and to keep checking to make sure the tanks are balanced according to the manufacturer’s procedure.

Welcome to Av.SE.  Your answer is correct & it does answer the question, but you're getting down-votes because it doesn't really add to the other answer here, doesn't explain "why", and doesn't reference any sort of authoritative source. So it ends up looking like "somebody on the internet said..." which isn't particularly useful. Can you add a reference or an explanation "why" we do it this way? That would improve the answer considerably.


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quadcopters — should the propellers/motors all be at the same height
In quadcopters (or in any other drones), must the propellers/motors all be at the same height, or can there be symmetric height differences (like for eg, two of the motors at the "back" are elevated just a little like by say 5cm or even less... Is that going to make a significant trouble in the drone performance?

Questions like these are the exact reason for SE.Drone

Tandem rotor helicopters have the aft rotor elevated to keep it clear of the front rotor's downwash and improve performance in forward flight (at the cost of reduced performance in rearward flight).
Most small quadcopters achieve substantial pitch angles in forward flight, often in excess of 30 degrees, which keeps the aft rotors out of the downwash anyway. Therefore, your proposal changes nothing in hover and fast forward flight, improves performance somewhat in slow forward flight and reduces performance in rearward flight.

what if opposite propellers are at the same levels, but the diagonal pairs are at different levels?

@space - If one rotor is downstream (aft and below) of another, that rotor will have reduced performance.

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sort: votes


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Why hasn't anyone tried making two propellers facing upwards and the other two facing downwards?
Within a quadcopter, I know usually it's all the propellers facing upwards or all of them facing downwards. Why hasn't anyone tried making two propellers facing upwards and the other two facing downwards? Do some forces cancel out and renders that method ineffective?

Do you mean two propellers each attached to one shaft, or still four propellers each attached to its own shaft? (A simple illustration of what you have in mind might help.) What do you feel the benefits of this construction would be?

Because it increases construction and maintenance cost and gives no benefit? I am guessing nobody simply has been able to think of a reason to do it.

People have done that.
And it works.
They do it because it's one of the logical configuration for a quadcopter based VTOL plane.
Example:

Who makes that plane? I want one!

@RobertDiGiovanni It's DIY. The plane itself is a Hobbyking Bixler (hobbyking.com). Probably using a KK2 or Ardupilot or some other open-source controller with the software modified to support VTOL transition. If you go to the VTOL forum on RCGroups (rcgroups.com) you can probably find the project or something similar

Brilliant, can not say enough praise.  A full scale could feather one set and/or fly with much lower power draw.

The elegance and simplicity of a quad-copter design lies in how it achieves control in all axis with simple variation of engine power (and thus speed) using just simple fixed-pitch rotors:

varying power to the forward pair of rotors compared to the aft pair controls pitch
varying power to the left pair of rotors compared to the right pair control roll
varying power to the clockwise spinning pair compared to the counter-clockwise spinning pair control yaw (this means the rotors placed diagonally spin in the same direction)

These commands work independently if all the rotors generate the same lift and torque for the same power. They also need to be laid out in a square or not too narrow rectangle. And the easiest way to achieve that is if both the rotors and their mounts are all identical (except for mirroring for counter-rotation).

Does not really matter if they are facing up or down, it matters how they are pitched and what direction they are spinning. All 4 with the same orientation is a good common sense approach to a quadcopter, though you can have opposite spinning pairs to cancel torque.  This is common on twin engine aircraft.
How ever, notice in a helicopter the tail rotor is tilted 90 degrees out of plane with the main rotor to control yaw torque from the main rotor.
And there is nothing to say you can not have more than 4 rotors, if you find an application where more is better.  Really a matter of matching (any) available technology that can be practically applied to the idea.
There might be slight differences in performance up vs down due to air circulation interferences near the prop, so the safer bet is to keep the design consistant, but 2 up and 2 down is not outside the realm of possible.

On a quad-copter you must have couter-rotating pairs. Otherwise you'd need an anti-torque rotor and that would no longer be the quad-copter layout.

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Commercial operation certificate requirement outside air transportation
My question is about whether I need a part 121/125/135 certificate to perform commercial operation that is not air transportation.
14 CFR 119.1 (a) says that part 119 applies to air carrier or commercial operator. According to the definitions in 1.1, both involve "carriage of persons or property".
So what happens, for example, if I post an ad to fly over your house in figure eight for you to enjoy for $200 an hour. In this case I'm getting compensated, I'm "holding out" for this operation, but it does not involve carriage of persons or property. If I'm reading the regs correctly, part 119 does not apply so I don't need a part 121/125/135 certificate. (I explicitly picked an activity not covered by 119.1 (e) to avoid potential confusion.) Do I just need a commercial pilot license and can I operate under part 91?

I would pay for this service! (Probably not $200, though. :))


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Why does a rectangular wing stall first at the root?
Why does a rectangular wing stall first at the root? as opposed to tapering of the wing which causes the stall at the tip and moving inboard?

The tip vortex (caused by higher pressure air from below the tip moving into the low pressure region on top by rolling around the tip) reduces the effective angle of attack at the tip and keeps the boundary layer attached longer as the stall progresses.  The tip vortex has this effect with any wing shape, but shapes that reduce drag by shrinking the tip vortex will gain less from it, while a wing with less chord will tend to stall at lower AoA -- causing highly tapered wings to stall at the tip first.
I say "gain" because a tip stall is usually considered bad -- it can cause a spin entry, where a root stall with the tips still flying leads to a straight-ahead stall, other factors equal.

@ymb1 Washout has the same effect and purpose -- and works on wings with substantial taper or sweep.  A "plank" style wing, zero taper and zero sweep, doesn't really need it (or needs a good bit less).

A rectangular wing is a cheap way to get washout.  Twisting the wing so that the tip has lower AoA is difficult to manufacture and costs more.  A rectangular wing gets washout by using the vortex however there is the cost of more induced drag.  If you want your plane to be spin resistant you need washout.


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What is the relationship between the Center of Pressure and flap deployment?
I have been reading regarding the Center of Pressure of an airfoil (Cp) and the relationship with angle of attack.
What is a good way to explain in layman's terms why the pressure changes in such a way as to move the Cp forward as AOA is increased?
Also, in a flap deployment situation; the Cp moves aft. My reasoning here would be because now the aft portion of the airfoil is now at an increased camber, meaning that you have more lift production towards the aft of the airfoil, the Cp (average of the lift production across the wing) would be moving aft. Is this correct?

Your first question is answered in the second half of this answer. I admit it is hard to find. Your assumption for the second question is correct; flaps will change the camber and thus the camber-related pressure distribution.


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Why is an elliptical wing planform so aerodynamically efficient in layman's terms?
Below is part of an answer to the post 'For the elliptical wing, what is elliptical, and why is drag regularly distributed?':

On the untwisted elliptical wing the local lift coefficient is constant over span, and changes in angle of attack over the linear range will change the lift coefficient equally everywhere. This, when combined with the elliptical chord distribution over span, means that the circulation distribution will stay elliptical over the linear angle of attack range. This is the special characteristic of an elliptical wing: While any wing can have an elliptical circulation distribution at one angle of attack (given the right twist distribution), the elliptical wing will keep that elliptical circulation distribution over the whole operating range.

My question is: how can the wing tips of an elliptical wing be considered to produce the same amount of lift as the root (see emphasis above)?

I believe that you misunderstood the answer that was posted.  The local lift coefficient is different from total lift.  In fact the question itself states the lift and drag decrease from root to tip.  If you include the previous two lines by Peter you will get a better context:

It is both the planform and the circulation distribution. Note that circulation is not lift coefficient but bound vortex intensity. You can interpret it as local lift coefficient times local chord.

The amount of circulation created is related to the total lift.  See the Kutta-Joukowski theorem.  Peter says the circulation distribution is elliptical along the span (decreasing towards the tip) so the lift is not constant along the span of the wing.
A more Layman's way of understanding this: The lift decreases at the tip since the chord length decreases but the local lift coefficient remains the same.  Since the local lift coefficient is related to the local drag coefficient this also means that the local drag coefficient remains the same along the span.  But again because the chord length decreases along the span the total drag actually reduces at the tip of the wing.

It makes much more sense now, you are correct in the assumption I misunderstood the original answer. Many Thanks.


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Is flying as a post-retirement occupation possible?
Is it possible for a non-professional pilot to pursue a post-retirement career as a professional pilot? Of course I'm not talking about airline jobs but what about bush piloting or missionary/non-profit work?

Can you pursue it, sure. Will you have the time to get to the point where flying is a viable second career for you? That depends on more factors than an answer here can cover. It would be a pretty unconventional path, certainly. Most people who pursue a career in flying start in their twenties, and many endure several years of pretty meager wages with a view to a few decades in the airline industry making it a worthwhile sacrifice. Starting in your 50's or 60's, your time horizon to make it financially worthwhile is far shorter.

@RalphJ From the tone of his question, I don't think he's worried about the financials. More - could I actually get some work?

@Cloud I would even say - can I get a work that will actually pay me for doing what I love (being a pilot) rather than asking me to pay for it

@Ister He asks about unpaid volunteer work as a hint

As you mentioned missionary, there's MAF, but I'm not sure about their age restrictions: https://www.maf.org/serve/serve-overseas/aviation

It depends on your jurisdiction. A lot of places mandate a retirement age of 65 for scheduled commercial pilots (part 121 here in the USA). You are allowed to fly 91 and 135 (charter) operations over 65. While you state you don't want to fly for the airlines, it's worth noting that this retirement age causes a large influx of older out of work pilots many of whom still want to fly. As such there are plenty of high experience older pilots that are primed for flying 135 operations. These may be the people you are competing against for a job which as a low time, potentially inexperienced pilot would be an issue at the bigger outfits. In reality it will depend on a bit on if you are getting into it to make a living or just to fly and bring home some extra cash.
With that in mind there are plenty of commercial operations that seem to always be looking for people:

All of the flight schools near me seem to constantly be on the hunt for instructors. If you are willing to go that route I would think you should have no issue finding work.
There is a small charter op run out of the FBO I fly from. They run 3 King Airs and seem to always be short of qualified pilots. A lot of people use these as time building/jump off points so turn over is high. Presenting yourself as someone willing to stick around, has no interest of going on to the airlines, looks pretty good. There are lots of these smaller outfits, generally at local airports.
If your willing to move to somewhere potentially far far away mission flying is an option. This episode of Omega Tau covers it in great detail and is worth the listen.
If you are looking to spice up retirement, ferry flying can be high pay for the right situations.
Depending on where you are there is a fair bit of Ag Flying you can do.
If you are in or near a beach area and like to fly low and slow, towing banners is always an option.

It all depends on the kind of flying you’re doing but test you can continue to work professionally as a pilot doing Part 91 or similar work.  I have a friend who just retired from UAL as a 787 captain and he now flies an Embraer Phenom 100 on the side.

Thanks for your insight. It's much appreciated. After many years of working to "bring home the bacon", I'd like to spend my retirement doing something I enjoy.


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Quadcopter: would putting 3 blades on the same level and the other blade below the first three work?
Most quadcopters have their propeller blades all on the same level, or in some cases, a pair of blades placed a little above the other pair.
However, I was wondering if three propellers of a quadcopter can be put on the same level and the other propeller a little lower. Would it just make the quadcopter inefficient?

To clarify, are you asking about a layout where a single propeller is in the center of the drone, and three other propellers are sticking out in a triangle shape?

There are designs with overlapping blades. It's all a tradeoff on space, structure, clean aerodynamics, efficient aerodynamics, etc. This is the Malloy Hoverbike, for example.

So you can definitely have them at different levels, and you can even definitely have them overlap. If the 3 rotors' flow is not involved substantially with the 4th rotor's (not too big a difference in height), it should be no big deal.

The OP specifically mentioned a pair of blades higher and a pair lower, which your Hoverbike example shows. He's asking about 3 of the 4 being at level 1 and the 4th being at level 2.

The OP didn't mention an overlapping set, which was the point of the picture (all kinds of configurations possible). Updated answer to address your comment.


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Why is A320's CG optimum position at 25% MAC?
I have read a320's optimum CG position is 25% MAC. But, the aerodynamic center is also at 25% of chord.
So since CG and aerodynamic center at the same position, there would be no moment arm of lift, so what will be the use of tail plane?
Also, as fuel burns,
the CG shifts rearward,
in that case aerodynamic center will be in front of center of gravity.
wouldn't that mean center of gravity is behind the center of pressure?
I have read center of gravity is ahead of center of pressure in airliners, so wouldn't this be a violation.
Please explain my two questions.,

Fly by wire baby~

Without a tail plane how would you control pitch?

I do not know much about the specifics of the A320 design, but I will try to come up with some general answers to your questions.

So since CG and aerodynamic center at the same position, there would be no moment arm of lift, so what will be the use of tail plane?

Even if it was the case in the A320 that the CG and AC coincide (I do not know), it does not mean that there is no moment from lifting forces. The aerodynamic center is not the point of zero moment, it is the point of constant moment within reasonable angles of attack. Hence there will be a moment from the lifting force. Assuming CG and AC coincide, the moment generated by lift will not change the total moment on the aircraft. Therefore a stabilizer is still required.

Also, as fuel burns, the CG shifts rearward, in that case aerodynamic
  center will be in front of center of gravity. wouldn't that mean
  center of gravity is behind the center of pressure?

Not necessarily. The center of pressure is usually for most angles of attack behind the aerodynamic center. Therefore, under the assumption that the CG and AC coincide, the CG can move aft of the AC and still be in front of the center of pressure.


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if center of pressure moves forward with angle of attack, then does it move forward of center of gravity also at a particular angle?
Consider an aircraft where-
Center of gravity lies forward of Center of Pressure in CRUISE.
Now as angle of attack increases,
CP moves forward,
then for particular angle, it must move forward of CG.
In that case, wouldn't longitudinal stability be lost?
As , now the CP would be making nose pitch up moment, the tail plane will also make nose pitch up moment ( as it acts downwards)

Longitudinal stability is related to Neutral Point (NP) rather than CP. Exactly because CP moves around, it is not a convenient reference point.
There is no strict requirement for the tail to create downforce for a stable airplane, even though this is a common arrangement.
It may well happen that CP may move ahead of CG and the aircraft remains stable. But it won't keep doing it indefinitely: at some point the wing will stall, CP will move back (this will depend on the wing plan form though), and NP will cease to exist as such. Stability analysis is difficult in such regimes.

if CP moves ahead of CG , and the aircraft is in stall, then in which direction will the tail create force?

As always in stall, it depends. Is the aircraft still in balance? Such question rarely arises (outside of aerobatics), given that the stalled condition is not meant for flight. But if it is, then by definition the tail must be creating positive lift. If not, there is nothing to talk about, it can be anything. That said, being stalled with CP ahead of CG is rather unusual situation, given that on most wings CP will move back after stall.

Assuming a "classical" planform with CG near the first 1/3 of the wing chord and a horizontal stabilizer in back of the aircraft, yes, the center of pressure of the wing could move forward of the CG when angle of attack increases, resulting in a pitch up force.
If you ever scratch build a wing it is fun to watch it tumble end over end if you try to "fly" it by it self.  The forward shift of CP is normal as AoA increases.  This is mainly due to breakdown  of lift producing airflow over the top back part of the wing that gets progressively worse until stall.  The wing will tumble over backwards over and over again.
What to do?  Notice the horizontal stabilizer also pitches up as AoA increases, and has a much longer lever arm to control pitch compared with the wing CP.  So you try to "right size" your H stab to exactly match the torque of the wing CP moving forward.  This helps keep the NET CP in the same spot.
Also, notice Hstabs are either delta shaped or "chunkier" lower aspect design that wings.  This means the wing stalls first! The H stab then helps pitch the nose down to restore proper air flow over the wing.  (No, Burt Rutan did not invent this, but it was applied very well to his canard designs).
Finally, even if the elevator trim is set to provide down force to the tail at cruise, as AoA increases, tail lift force will go from down to nuetral to up on its own, or the alert pilot will apply down elevator to avert the stall!


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Confusion about Aerodynamic center
suppose, i assume Aerodynamic Center lies at Center of gravity.  Then apparently moment by lift will be zero about CoG. But in reality Centre of Pressure may be behind aerodynamic center and still be producing a moment. So isn’t using aerodynamic center absurd?

Related

It's easy if we clarify your statement a bit.
When aerodynamicists say "moment by lift", this means the derivative of moment by lift. That is, the change of moment per change of lift, $\partial M/\partial L$, or shorthand $M_L$ (or their coefficients more commonly: $C_{M_{C_L}}$). Stability analysis (that's where the aerodynamic center is most useful) is all about derivatives rather than actual forces and moments: we are interested what changes when conditions change.
In this light, "moment by lift" will be zero, true, but "just moment" won't be.

This makes much more sense.

so can we say that neutral point is an aerodynamic center itself? because dM/dAlpha equals zero condition is satisfied here. isn't it ?

Yes, in normal usage, neutral point is the aerodynamic center of the whole aircraft. Sometimes the terms are used interchangeably when the contexts is clear.

thanks, can you also clarify one more thing for me.
do spoilers only decrease lift or , they increase drag also?
some say spoilers only decrease lift but on wiki it says it decreases lift and increases drag?

This is worth a separate question, but briefly, both, of course. It's just some people like to point out that the main point of spoilers is to decrease lift, which may feel counter-intuitive.

First of all, I am going to comment on your statement

Then apparently moment by lift will be zero about CoG.

This is not apparent and not necessarily true, as your next comment also implies you know about.

But in reality Centre of Pressure may be behind aerodynamic center and
  still be producing a moment. So isn’t using aerodynamic center absurd?

I cannot answer whether or not the aerodynamic center is absurd. I will instead give an explanation of why it is reasonable to define the aerodynamic center.
You are right in the fact that the center of pressure may be behind the aerodynamic center and hence produces a moment around the center of gravity which you assume equal to the aerodynamic center. In fact, a lifting force through the center of pressure produces a moment around any point not equal to itself, hence also the aerodynamic center.
Consider an airfoil at a given angle of attack and fix the airfoil in some arbitrary point p along the chord line. We can now (at least theoretically) determine the pressure distribution around the airfoil and use that to calculate the aerodynamic force from these pressure variations and the center of pressure. The force acting through the center of pressure gives rise to a moment around p proportional to the arm, i.e. the distance between the center of pressure and p.
If we change the angle of attack, the pressure distribution changes. Therefore, the aerodynamic force and the location of the center of pressure and the moment all change. If we had still fixed the wing at p, the size of the moment would most likely have changed by now.
We see that determining the aerodynamic behavior of an airfoil is very complicated if we use the center of pressure to analyze the forces because all variables (leverage, force and moment) change when we change the angle of attack.
We could have computed the moment about any point on the airfoil. Given a pressure distribution, the aerodynamic force will be the same, but the value of the moment depends on the point where that force is applied. It has been found both experimentally and theoretically that, if the aerodynamic force is applied at a location roughly 1/4 chord aft the leading edge on most low speed airfoils, the magnitude of the aerodynamic moment remains nearly constant with angle of attack. Engineers call the location where the aerodynamic moment remains constant the aerodynamic center of the airfoil. Using the aerodynamic center as the location where the aerodynamic force is applied eliminates the problem of the movement of the center of pressure with angle of attack in aerodynamic analysis. The aerodynamic center is important when determining the longitudinal stability of an aircraft.
You can think of the aerodynamic center as the point through which the part of lift, which is dependent on the angle of attack, acts.
A figure showing the constant aerodynamic center for angles of attack equal to 8, 4, 0 and -4 degrees. Notice the negative lift on the forward part of the wing at negative angles of attack compensating for the large leverage of the center of pressure (here denoted TC).

But if my CoG somehow coincides with Aerodynamic center, I multiply the lift produced by the distance to the center of gravity which is in this case is zero. So mathematically the answer is coming zero but in reality it is not. So isn't this a shortcoming?

The lift acts trough the center of pressure, not the aerodynamic center - see the illustration. Here TC is the center of pressure. There will be a moment UNLESS the center of pressure coincides with the center of gravity.


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When an aircraft is descending, does any lift act on it? If yes, how?
When descending , an aircraft never goes in free fall, even if negative angle of attack is large. Does that mean the net downward force is not gravity, but less than gravity, due to certain lift acting at all negative angle of attacks?

This question is unintelligible.  A large negative angle-of-attack would create a downward loop.  That is very possible.  If the wing is at the zero-lift angle-of-attack the aircraft would go into freefall.  That is also very possible.  Yet the language of the question suggests otherwise.  No on should be answering this question because it is posed it a manner implies things are true that are actually not true.

I really don't understand why so many people are answering this question, and why their answers are getting upvotes.  I though that the way this site works is that a question that contains contradictory or false statements-- statements not congruent with reality-- should be edited, not answered.

Re "an aircraft never goes in free fall", that's not strictly true.  See e.g. the "Vomit Comet" used for astronaut training: https://en.wikipedia.org/wiki/Reduced-gravity_aircraft  Can be done in other planes, too - even sailplanes.

Why does the concept of "negative angle-of-attack"  have anything whatsoever to do with a normal descent, such as a final approach to landing, etc?  And who says an aircraft "never goes into freefall"?  Nobody should be answering this question until it is edited to be less self-contradictory.

@quietflyer, the question is self-contradictory, because its author is confused about the proper definition of the terms. It should be answerable by clarifying what they actually mean (we would need some feedback from them though regarding whether the existing answers clarified anything and what remains to be unclear).

Okay. I would like to ask this,
Suppose an aircraft is flying at 5 degree angle of attack.
Now if a pilot applies forward pressure on control column, and the aircraft pitches down greatly. Then the aircraft has achieved negative angle of attack. Now the coefficient of lift becomes negative according to the graph "Coefficient of Lift versus AoA". Now what does this negative lift indicate?

@Sachin Are you aware of the difference between pitch angle and angle of attack? I based my answer on the suspicion that you do not, but from that comment I still can't tell. Negative lift would be where you push so hard on the yoke that your head would hit the ceiling if you hadn't buckled up; is that what you're asking about?

@SachinChaudhary just because you push forward on the stick, doesn’t mean you will get a negative angle of attack. It will certainly reduce but you’d have to push forward quite hard really quick to get negative angle of attack. Even then it isn’t forced to stay negative long.

You seem to be confusing angle of attack and pitch angle. The pitch refers the orientation of the aircraft: where is the nose pointing? The angle of attack is the angle of the wings relative to the incoming air stream. Both can take on any value independent of each other. For example, one could imagine an aircraft falling straight down, but wings level. The pitch angle is then zero, but the angle of attack is 90 degrees!
Similarly, the pitch angle can be negative for a continuous steady state descent (and often is for a steep descent). However, you correctly identify that an aircraft on approach generally does not go into a freefall. The lift acting on the airplane is still (approximately) the same. A positive angle of attack is thus still required. As a result, the aircraft will be descending even steeper than where the nose is pointing, resulting in a positive angle of attack (the angle between the pitch vector and the even steeper velocity vector).

The Newton's second law of motion tells us that acceleration of object equals sum of forces acting on it divided by its mass. In a steady descent, an aircraft is not accelerating, it is just flying with a constant velocity pointing obliquely downward. Therefore the sum of forces acting on it must be zero.
Now there are four main forces acting on an airplane:

(Image from How It Flies, chapter 4 Lift, Thrust, Weight, and Drag)
While drag is slanted slightly upward in descent, and thrust might be depending on the resulting pitch attitude, in normal descent the only force with large upward component is still lift, and therefore it must still be almost equal to weight to get the forces in balance.
The lift is only decreased when initiating descent, so the plane needs to accelerate downward in order to change direction. However usually it only decreases by maybe 10–20%. If it decreased more, you'd surely notice: if it decreases to zero, you'll feel weightless (e.g. in the Vomit Comet) and if it went negative, you'd raise from your seat as the aircraft would be accelerating downward faster than free fall, but you only have gravity pulling you down, so you'd lag behind it.
Besides the above linked, you may also want to look at the chapter 2  Angle of Attack Awareness and Angle of Attack Management of the abovementioned How It Flies book, for detailed treatment of the relationship between Angle of Attack and pitch, and any other section too; it does quite a good job explaining the physics related to flying.

I see different ways the lift is illustrated. Some say the lift is "the force that directly opposes the weight of an airplane", meaning the lift is opposite to gravity, i.e. perpendicular to the horizon. Some illustrate lift as perpendicular to the aircraft's chord line, so not necessarily perpendicular to the horizon (when the aircraft is pitched at an angle). In the diagram you show above, the lift is perpendicular to the wind direction. So confusing. Is there a universal definition, or people use the word "lift" loosely in different ways? I wish I know how to attach pictures here.

@gadfly, the universal definition is as in this picture: lift is the component perpendicular to relative wind, drag is the component along relative wind. Most images show level flight, so the distinction is not clear in them.

@gadfly, also the force generated by airfoil due to circulation is perpendicular to it, but that includes the induced drag and gets decomposed to lift, perpendicular to the relative wind, and the induced drag, parallel to the relative wind.

Thanks. So the so-called "lift induced drag" is really not induced by lift (which is perpendicular to the drag"), but it's induced by that resultant force (the hypotenuse of the right triangle, the two legs of which being the lift and the drag)? (a force can not induce another force perpendicular to itself)

@gadfly, the lift induced drag is induced by lift in the sense that any way of producing lift over finite span also produces this drag. It is not a component of lift, but the term “induced” would not be used for describing that anyway.

Thank you for making it much clearer to me. This picture above is good and more general in that the three directions---horizon, airfoil position, travel path (opposite of wind direction)---are all different. Many other sources have all three in the direction/plane.

My only minor issue with this is that the direction of travel isn't always opposite to the direction of the relative wind

@DaveGremlin, I believe “direction of travel” does not imply whether it is ground or air coordinates. Relative wind is, obviously, air coordinates, so if you take the former, they indeed may differ.

@Dave Gremlin My (total layman's) understanding is, relative wind is by definition the exact opposite of travel direction. I think this word 'wind' here does not refer to the wind in weather context, but that when you move fast in a direction, you feel the wind coming at you. Am I correct? (if so, I try to imagine a perhaps improbable case, when the natural wind blows in the same direction as the aircraft's travel, and wind speed is as high asthe aircraft's travel speed, then the relative speed between the aircraft and the air around it would be about 0, would there still be lift, then?)

@gadfly, relative wind is indeed the wind that you feel coming at you due to your motion, but that means you have to be moving relative to the air mass. If natural wind blows in the same direction as the aircraft's travel, the aircraft will maintain its speed relative to the air mass, because it needs that to generate lift, and it's speed relative to the ground will be sum of the aircraft airspeed and the wind speed.

The short answer is Yes.  In a "normal" descent, (i.e. wings level positive G) there will still be a positive AoA.  All that needs to be done to initiate a descent from a state of equilibrium is to reduce a little power and adjust pitch nose down enough to maintain airspeed.  This will initiate a downward acceleration, but once the rate of descent has steadied out the wings will still be producing lift equal to the weight of the aircraft that they are opposing.
If there is a large negative AoA as the question stated, (assuming wings level) the aircrew would experience negative G forces and the aircraft would accelerate downwards at 32 ft/sec squared, PLUS any additional downwards force produced by the wings as a result of the negative AoA.  Suffice it to say that this would be a rather brief transitory state.

Re "The wings are still producing lift, just a little less lift than the weight of the aircraft that they are opposing": no, unless the plane is accelerating downwards, the lift of the wings is still equal to the weight of the aircraft. Although maybe a small part of the lift is due to the downward speed, so one might call that part an upward drag force.

Your answer makes sense, because the aircraft reaches a steady state.  (not accelerating)  However, I would seem to me that that forces cannot be equal.
Consider the classic diagram of an aircraft turning at 60deg AOB.  Load factor must be increased to 2Gs so that the vertical component equals the weight or the aircraft will descend in the turn.
How do you account for this?  If it in balanced flight and holding a constant altitude, and then is descending at a constant airspeed, something has changed.  What has changed?

@MichaelHall What changed is that during some period between the steady state of flying straight-and-level and the steady state of descending at a constant rate, the aircraft briefly produced less lift than its weight, causing it to accelerate downward. Once the desired vertical speed was reached, the lift was returned to equal the weight in order to maintain a constant rate of descent. On the flip side, as you slow the descent, the aircraft will actually be producing more lift than its weight, even though it's still descending, as it would be accelerating upwards.

@MichaelHall More to the point, acceleration is directly proportional to net force. Current velocity is irrelevant to that relationship.

@reirab, thanks - good explanation.  I edited my answer above to reflect this truth.

@MichaelHall An aircraft that is turning is accelerating in the horizontal direction, towards the center of the turning radius. The "2G force" is perpendicular to the banked aircraft, not perpendicular to the ground, and can be split into 1G acting vertically downwards (to balance the weight of the aircraft) and a component acting horizontally, which causes the acceleration which makes the plane fly in a horizontal circle - i.e. "turn". Note, there is also a smaller horizontal force from the rudder, which rotates the plane as it turns, to keep the nose pointing in the right direction.

@Marc, no, you have your definitions wrong. Lift acts perpendicular to the airflow, not "up" (this answer has a correct diagram). So in a descent the wing is relieved a bit by that "upward drag force" (and possibly thrust), and this force is not part of lift. But yes, all forces are in balance, of course, and this answer makes a clear point about that.

@alephzero, thanks - I know how airplanes turn, I was just trying to clarify a point about the balance of forces in a descent.

During a steady (non-accelerating) descent, the sum of the vertical components of lift and drag equal the weight of the aircraft. The total lift (aerodynamic force perpendicular to the aircraft's path with respect to the air) will be reduced. Thrust and drag may may also be reduced depending on the air speed.

In an aircraft that is descending at a constant rate, not only is there lift acting on it, but that lift is equal in magnitude to the aircraft's weight. If the lift were less than the weight of the aircraft, then the aircraft would be accelerating downwards.
The only times that lift is not equal to weight are the times when the aircraft is accelerating upwards or downwards. That is, when the aircraft changing its rate of climb or descent. Any time when the rate of climb or descent is constant, the lift is equal to the entire weight of the aircraft. As Jan Hudec's answer mentions, this is a necessary consequence of Newton's Second Law.
As others have mentioned, remember that pitch angle and angle-of-attack are two totally different things. Pitch angle (angle of the nose relative to the horizon) can be quite negative while still maintaining a positive angle-of-attack. During a normal descent, angle-of-attack is always positive.
Note: For purposes of this answer, I'm defining 'lift' to be "the sum of all forces acting on the aircraft in the opposite direction of gravity" in order to keep things more simple.

Any particular reason for the downvote?

Unless you’re flying some sort of unlimited aerobatic airplane capable of such rapid pitch changes that you can achieve a negative angle of attack, your angle of attack will always remain positive as the relative wind changes in direction when transitioning from straight and level to a descent.  A descending airplane is subject to the same forces and loads as any other airplane is.  However in descent, gravity does provide a component of its total which acts in the direction of other thrust forces, propelling the aircraft forward (this is also how a gilder works) as the potential energy of the airplane is siphoned off and converted into work to counteract aerodynamic drag forces.

Gravity always acts downwards. How can gravity provide a component?

Orient your frame of reference with the aircraft’s axes and you’ll understand why.

When descending, an aircraft never goes in free fall,

Not strictly true. Freefall just means falling at the same rate you would accelerate under gravity. The passengers on the vomit comet experience free fall even after the aircraft has peaked and has started to descend. Aerobatic maneuvers can often result in zero or even negative G forces.

...even if negative angle of attack is large.

Angle of attack is not actually relevant here. If the wings fell off your aircraft, you would still descend and you still wouldn't experience sustained freefall.

Does that mean the net downward force is not gravity, but less than gravity, due to certain lift acting at all negative angle of attacks?

Yes and no. The upward forces acting on any object descending through the air can include drag from air resistance, lift from aerodynamic effects, and thrust from engines etc. The downward forces obviously include gravity, but could also include thrust from engines and negative lift from aerodynamic surfaces. So while lift (positive or negative) from angle of attack is a factor, it is only one of many.


text:
How does increasing the rate of pulse pairs in DME 'search' mode help?
From How does an airborne DME unit differentiate itself?

The higher rate in SEARCH [150 PP/S] gives it more opportunity to detect the replies as it has no knowledge of range in search and thus has to scan for the entire range. Once it has detected the ground station replies by finding a correlated sequence of pulse pairs, it can switch to TRACK mode [30 PP/S].

How does increasing the pulse pairs per second (PP/S) here matter at all, since there will be replies anyway at the slower rate?

Does this answer your question? How does an airborne DME unit differentiate itself?


text:
Why For A320, On ground, reopening of the pack valve is delayed for 30-60 seconds
I've read in FCOM that, during engine start reopening of the pack flow valve is delayed for 30-60 s to allow second engine start? But isn't it that the pack flow remains closed only for the side the master switch is on? So how does that evgen affect second engine?

But isn't it that the pack flow remains closed only for the side the master switch is on?

This is correct, when the crossbleed valve is closed only one pack valve is closed and reopened with a delay. When the crossbleed valve is open, then both pack valves are closed and both are reopened with a delay.

Source: A320 FCOM

The documentation, which could be improved, must be interpreted this way:

Crossbleed valve closed:
In this case only one pack is connected to the starting engine. Only one pack vane is closed, according to bullet 1. It is reopened when the conditions are no more the ones in bullet 2.
When the second engine is started, the same sequence applies on the other pack vane.

Crossbleed valve open:
In this case the two packs are connected to both engines. When the first engine is started both pack valves close, according to note on bullet 2. The reopening is delayed on both valves, according to bullet 3. This delay anticipates the start of the second engine which would need to close them again.


text:
What RA will be generated if intruder traffic is at our level in TCAS II?
If the other traffic is at the same level, and neither climbing or descending. Will the RA say "CLIMB" or will it say "DESCEND"? How will it be decided and what are the possibilities?
Assume:

Mode S is not available
Only one aircraft has TCAS

Will it be random?

@ymb1 that concept makes sense to me, if it is the case that the "coordination" mentioned is not the same as the "negotiation" that is apparently only available in later versions of TCAS. I'm not clear on that. Also, if the two aircraft are truly level at the exact same altitude, how does the more-authoritative aircraft (lower Mode S) determine what the solution is? Does it, for example, decide to dive (an easier maneuver) and force the other one to climb? I could imagine that being the case but I do not know.

The TCAS algorithms are specified in RTCA/DO-185B (not freely available).
In case of an encounter at same altitude with a non TCAS equipped aircraft, assuming that neither climb nor descent are inhibited, the TCAS will produce a descent advisory (part 2, sec. 2.2.4.1.2.1.5 - sense selection).
This is a consequence of the somewhat subtle signs used in the specification (<= for descend, > for climb).

A non mode-S intruder will broadcast altitude with a 100ft resolution, and a TCAS equipped aircraft will normally use barometric altitude with 25 ft resolution, so the two aircraft do not need to be at exactly the same altitude - in the worst case, a difference of up to 62 ft will be interpreted as "same altitude".

There are also provisions for climb and/or descent inhibits. In particular descend RAs are inhibited below 1000ft radio altitude.
Climb RAs are inhibited when the aircraft performance are not compatible with a 1500 fpm climb (for example, one engine out). The exact mechanism through which this information is provided to the TCAS computer is aircraft dependent.
In your case of same-altitude encounter, if descend RAs are inhibited, a climb RA will be generated.

EDIT
As requested, more clarification.
A simplified description of RTCA-DO185B initial sense selection follows. This description assumes that the intruder is not TCAS equipped, and that the firmness (confidence in intruder altitude and position) is sufficiently high.
First, choose the direction (climb or descent) that will produce the highest separation at CPA (see below).
If equal, select descent (see 3.52 Noncrossing_Biased_Climb and 3.53 Noncrossing_Biased_Descent).
Denote this direction with D

If a RA in the D direction does not result in a crossing (see below), then issue an RA in the D direction.
Otherwise, if the opposite direction still provides at least ALIM separation, then issue an RA in the direction opposite to D.
Otherwise, issue an RA in the D direction.

Separation: The TCAS algorithm models a RA response as follows:

5 seconds of reaction time (pilot maintains current airspeed and rate-of-climb)
A segment with a constant acceleration of 8 ft/s^2 in the direction of the RA
A final segment at constant rate of climb (usually + or - 1500 fpm).

To select the sense, the TCAS algorithm considers:

Distance from intruder
Time derivative of distance from intruder (closure rate)
Time derivative of the closure rate
Ownship altitude and rate of climb
Intruder altitude and rate of climb

With these 5 parameters, the vertical separation at the CPA (Closest Point of Approach) can be computed.
In this scenario, the relative altitude is zero, and the rate of climb of both intruder and ownship are 0. Due to the symmetry of the model, the separation will be the same for both climb and descent.

Crossing
Crossing is defined as either:

A climb RA, with ownship at least 100 ft below the intruder
A descent RA, with ownship at least 100 ft above the intruder

In this scenario, ownship and intruder altitude are equal, so neither direction will result in a crossing RA.

Can you provide us more details on that equation? I think it's strange that TCAS II considers only the altitude as parameter (see my answer and the NASA document)

And this FAA document: https://www.faa.gov/documentlibrary/media/advisory_circular/tcas%20ii%20v7.1%20intro%20booklet.pdf

It is not possible to give an answer, because the TCAS II RA algorithm is not so simple. It isn't like "if A is 50 feet above B then A climb and B descent", it consider many different parameters to take a decision. Under certain conditions, it is possible that even if our aircraft is below the intruder, the RA can instruct us to climb.
This paper from NASA describes the equations on which TA and RA TCAS II algorithms are implemented and states:

The sense of an RA is computed based on the direction for the ownship maneuver that provides a greater
vertical separation, with a bias towards the non-crossing direction.

This depends on both aircraft TAS, vertical speeds, horizontal separation, and many other parameters. The final equation is the following one (I highlighted the return values):

The comparison highlighted in blue tells us that in the rare case that performing a climb or a descent will give exactly the same vertical separation values, then CLIMB is performed.
See also Page 29 in TCAS II FAA booklet.

@ymb1 I don't have access to DO-185B, but I suspect that is for TCAS I and not TCAS II

@ymb1 I feel it as strange that depends only on altitude. TCAS II should consider also other parameters to decide the direction, not only the altitude

NASA document also agrees with this FAA document: https://www.faa.gov/documentlibrary/media/advisory_circular/tcas%20ii%20v7.1%20intro%20booklet.pdf

@ocirocir It does depend on many parameters, however OP is asking specifically for a "symmetric" conditions, with both intruders at the exact same constant altitude. In this particular case, both a climb and a descent will produce exactly the same value (own_alt_at in your answer),

@sbabbi but it depends also on the speed and the horizontal position... Also the NASA paper seems to provide the opposite answer...

thanks for the useful information, but which one is correct? @sbabbi it's conflicting between the two documents

The original poster's question assumes that mode S is not available.  The original poster did not stipulate that only one aircraft lacks mode S, therefore, neither aircraft has mode S.
Mode S is necessary for TCAS II, and therefore one cannot have a resolution advisory without a mode S transponder.
Under the conditions stipulated, no mode S and only one aircraft with TCAS, that aircraft can only have TCAS I, which will provide a traffic advisory, but not a resolution advisory.  With no mode S and no RA, there will be no direction given for a climb or descend, only a traffic alert.  It will be up to one or both pilots to see and avoid and to take necessary action to avoid a collision.

Only Mode C or above is required for RA generation in TCAS II . Please make sure you're current with the information before answering.

What are you talking about?  Mode S is a basic requirement for TCAS II.  TCAS II is necessary for a resolution advisory.  If neither aircraft has a mode S transponder, there is no TCAS II possible, and there is no resolution advisory possible.  Please educate yourself on TCAS II, mode S, and resolution advisories.  https://www.faa.gov/documentLibrary/media/Advisory_Circular/20-151B.pdf

Yes. Thank you.


text:
What would be VSI indications of an aircraft sideslipping to a blocked static vent while descending?
The aircraft has two static vents, each on left and right side of fuselage. Left static port is blocked and aircraft is sideslipping to left at an INCREASING rate while descending. Would VSI overread or underread?

Why do you think a static port would be "blocked"? It's a static port, it doesn't care about the attitude of the aircraft.

@JuanJimenez a static port can be blocked any time and there have been cases, due to dust etc

You are not talking about a static port being blocked by foreign matter, you are talking about sideslipping. No static port can be blocked due to sideslipping.

I'm talking about a case where a port is blocked and an aircraft is sideslipping. I've never said the port is blocked due to sideslipping.

The answer is still the same, if there are two static ports and one is blocked the other one still won't care if the aircraft is side slipping and neither will the VSI. I suggest you edit your question to properly state what you are asking.


text:
What would be the ASI readings incase of pitot leakage?
If there is a leak in the pitot tube in the a/c, what would be the readings for a pressurized and unpressurised aircraft?
I've read that if there is leakage in pitot tube within the structure of a/c, so for non-pressurised a/c it will under read because air will escape from the tube to the cabin, as cabin pressure would be less than pitot pressure?
Is it correct?

@mins, it seems quite clear to me. In a non-electric airspeed indicator there is a pipe from the pitot probe propagating the total pressure all the way to the instrument, and similarly from the static port propagating the static pressure, so the instrument can purely mechanically show the difference. So these pipes run inside the cabin (behind the instrument panel) and can be cracked there, affecting the readings.

@JanHudec: Ok, pitot tube vs. pitot tube pipe, it's the pipe which is leaking.

The underlying principle of how an airspeed indicator works is to display the difference between pitot (ram air) pressure and static (outside air) pressure.  The larger the difference, the larger the airspeed needle swings.
The instrument will convert that pressure difference into knots indicated and display an indicated airspeed.
For a 1 psid difference between pitot and static pressure, the airspeed will read approximately 145 KIAS.  At 2 PSI difference, approximately 205 KIAS.
This background is important to analyze your question.

If the pitot tube cracks in an unpressurized aircraft, the airspeed will show a small difference between RAM AIR and static air.  The difference will be too small to show anything on the airspeed and it will read 0 KIAS.
If the pitot tube cracks inside a pressurized aircraft, the airspeed will have a large difference between RAM air (cabin pressure) and static air.  The airspeed will be pegged at max airspeed.  As you descend, the airspeed will move to zero when you are unpressurized before landing.

In the pressurized airplanes I fly, a 9.2 PSI difference is standard.  That is quite a bit higher than what is listed above for 205 KIAS.

Even if there is a leak in the pipe, the ambient air is still flowing inside it and it still senses the pressure, however it is different. The ASI does NOT read zero. Incorrect.

@SachinChaudhary, it depends on how big the crack is. If it's large enough, it will read 0 (it can't display less than around 30 knots). But you are right that it may not read zero, it may read any value less then actual depending on how big the leak is (or any value more if the airplane is pressurized).

I would agree with these comments.


text:
Why are European low-cost carriers not running long haul flights to destinations like the USA, Asia, or the Middle East?
Does this benefit them somehow, by only travelling within Europe?

When you look at low cost carriers globally, the majority stick to short-haul. Doing long haul at low cost is very difficult.

well here are a few reasons off the top of my head.
LCC aircraft carry maximum payload so the galleys are not too big. Flying long-haul requires bigger galleys and possibly more toilets. This wastes space which on aircraft configured for short-haul is used to carry pax. In a nutshell a LCC aircraft can carry more pax.
The average sector lengths are shorter so more turns in a day. Before LCCs you would typically see a 737 or A320 turnaround take around 45 minutes. LCCs can do it around 25 minutes. That is a savings of 20 minutes. If you fly six sectors a day that is additional 120 minutes or 2 hours more that the aircraft can spend flying to make money. On long haul, you can't cut the turnovers much as cleaning and loading catering and cargo takes times.
On long haul you carry more fuel (in fact you carry more fuel to carry the fuel you need) so costs are higher.
The smaller aircraft allows improved frequency .. ie better to fly a small aircraft 6 times a day than to fly a widebody twice a day. This increased frequencies make it attractive to customers.
There are a lot more but these are the major ones..

sticking to within the shengen zone also simplifies customs regulations.

Some are, Norwegian Air runs a flight from Stewart airport in New York to Dublin Ireland and back.
One the benefits that LCC's get is from buying only one airframe for their fleet, which can, depending on where they are based limit their range as many fly A320's or 737's.
The benefit is not within Europe per say but with short, fast turn around, flights that benefit from the general implementation of the hub and spoke system. There are also LCC's domestically here in the USA like Spirit and plenty in Asia as well.


text:
Remote access to Area Forecast Discussion
Is it possible to access Area Forecast Discussion via SMS?  I fly to remote areas frequently (Wilderness), and after several days, it is almost impossible to get a good weather forecast, (one you're willing to bet your life on).  I wish I could get the Area Forecast Discussion via SMS over my inReach.

Thanks for the advice re. XM Wx.  I have used that, but I would like a longer forecast horizon - like a week. Here's what I've learned - you can download an FTP file of the Area Forecast Discussion, but the NWS system just responds to the email address that originated the query.  InReach doesn't allow replies to the initiating email address - even though it came from your device.  If they did, this would work.  I sent them an email last week but haven't heard back.

I have done something similar, and it might be possible to do it for the FA discussion. Essentially, I had a Google Sheet that would pull a table of data from a website, and in conjunction with IFTTT it would send a text message to a pre-defined number with the contents of that data table.  Since the avwx.gov page has a specific URL for each area discussion, this same concept would probably work.  Every 6 hours, the FA is updated, and you can time it so that the new discussion is texted to you at that interval.

But I'm confused: you said that you were looking for a product that had a longer forecast horizon of about a week... the FAs are only good for about an 18hr period.  ??

I have an old Garmin 496 & XM which works great for this. XM lets you suspend your subscription in winter if you want to. Tom M

You may want to look into something like XM WX Data which has pretty good remote satellite coverage for getting weather data when and where you need it. You may also want to consider installing an HF radio in your plane to potentially communicate with other ground stations in your area. Sometimes it boils down to getting in the air and calling other stations as discussed here.


text:
Landing gear sequence

{With reference to this video}
Is there any particular reason that land gear retract half way then 2nd gear retract half way and then proceed to retract completely.
Is it related to center of gravity (cog)?
2nd thing I noticed , Tyre boogie is tilted , is there any reason for it ?whats the mechanism behind it? I believed that aircraft land at particular angle , so that all wheel may contact ground at once .is it correct?

And this question probably answers your second point.

This cover all's thanks


text:
Is there a serious effort underway to move from a highways-in-the-sky approach to a mesh-network approach?
As far as I'm aware, there are certain "motorways" or corridors which air traffic uses, rather than navigating more directly from origin to destination.
I remember listening to a documentary which mentioned that in the early days of the standardisation of flight, the navigation system was being decided. The Americans put forward the "motorways" system because they already used intermittent beacons to guide aeroplanes across the country, which naturally form a kind of road system analogue.
While the British (or Europeans) put forward a direct point-to-point system.
Anyway, the US won, and everything was hunky-dory until we get to today with our extremely crowded skies. So the documentary went on to say that investigations were underway into a new system in which each plane communicates with the planes around it continuously, forming and reforming impromptu mesh networks to avoid collisions. In this way, planes can fly more direct routes and potentially save time and fuel.
My question is whether or not this suggestion is feasible, is it being investigated, and is there a push to move towards it at a level which could affect the international air community?

Possible duplicate of Why do RNAV airways exist?

@StephenSprunk Thanks for supplying a term I can use in subsequent searches, but I think my question is more along the lines of "Is modern technology making RNAV airways obsolete?"

Ironically, the U.S. nowadays is far closer to a mesh-network approach, with aircraft going pretty close to point-to-point, than Europe, where aircraft are compressed onto a relatively few, congested routes.

Currently SESAR (Single European Sky ATM Research) works on implementing a similar system in the European Union.
As stated on the website :

SESAR’s vision builds on the notion of trajectory-based operations’ [...] meaning that aircraft can fly their preferred trajectories without being constrained by airspace configurations.

The aim is to improve capacity in the European sky and be more flexible among the actors too (like temporary closed airspace for military training, handle volcanic eruptions and disruptions of the airspace).
SESAR has different projects to address the needs of imporving Air Traffic Management in Europe that you can find on their website.
Edit
There are current concrete solutions from SESAR and EUROCONTROL (the European organisation for ATM in Europe) to make this happen at the moment in the European airspace, such as the Free Route Airspace with set target of implementation (that we could consider "serious effort") :

By the end of 2017, 51 ACCs have either fully or partially implemented Free Route Airspace operations [...] By the end of 2019, most European airspace is expected to have implemented FRA, with all airspace having this type of operations by 2021/2022.

In the US, there has been an ongoing effort to accomplish accommodating direct routes as much as possible since the Free Flight initiative in 2000, and the National Preferred Route plan, beginning in 2008.
There are inherent problems with any such plan. Here are the trickiest ones -

Major airports must have a usable plan for flowing traffic to and from the field. This means traffic must be "gated" on arrival and departure routes (trust me, I've worked flows that had both, due to weather, and you, the user, DO NOT want that :) )
Radar and communications constraints. This applies mostly to "low and slow" aircraft, but a GNSS/GPS direct route cannot have a segment longer than 500NM (at time of clearance) for non-radar constraints, and fixes on the route must be displayed on low/high IFR charts.

I'll address the first item. If your airline files KSFO..KJFK, they will NOT get that route. KSFO has departure procedures and standard instrument departures (SID) that ensure that your flight will not be climbing out in the opposite direction of arriving aircraft.Your aircraft will be given a SID to comply with traffic and airspace constraints.
Now, you're 150 miles east of KSFO, and an Oakland ARTCC controller gets an alert that your current route will take you through the Utah Training and Testing Rage (UTTR)", which is reserved military airspace owned by Salt Lake ARTCC. Well, that's not good, so you'll get a routing of "BVL..TCH..KJFK" to take you through the area safely. On through Salt Lake's airspace to Denver ARTCC, where, just east of the Rock Springs, Wyoming area, you'll get a flow control route (ZDV does lots of these) for the preferred routing into KJFK. This will include airways (those highways-in-the-sky) that meet restrictions for Minneapolis, Chicago, Indianapolis and Cleveland ARTCCs, who all have their own traffic flow restraints.
Realistically, however, your airline will have access to this routing information before filing the clearance, and will file accordingly.
Departing a major airport and flying directly to another major airport over a long distance isn't realistic, unless it's a redeye flight, and you get lucky.
At least, not in the US.


text:
What does it mean to fly a procedure turn?
So, I'm flying IFR to Newport (KONP) for the ILS RWY 16. There's an aircraft on the approach, so I'm cleared direct BUHRS at 5000' to enter the hold as published.
I get to BUHRS, I turn outbound on the hold, and am "cleared for the approach". (Note: Not "straight-in").
So, what is my next step?
AIM 5-4-9 says I'm required to fly the procedure turn, but I can't find a definition of what that actually means (i.e. do I have to overfly the waypoint to "start" the procedure turn, or does it just mean remain in the protected area?). The options seem to be:
1) Turn back to BUHRS, overfly BUHRS, turn right (ending up exactly back where I am right now), intercept the localizer outbound, descend to 2600.
2) Intercept the localizer outbound, descend to 2600.
If #2 is correct, then there's an additional point of consideration. 5-4-9a1 says I have complete discretion in how I fly my procedure turn. Certainly turning immediately inbound fits within that discretion (instead of turn right, intercept outbound, turn left, turn left, turn right, intercept inbound). Or, what about continuing to fly outbound by heading for a minute then turn inbound and intercept the localizer?
A simplistic reading of the AIM seems to indicate #1 is correct. I asked AOPA, they say #2 is correct.
Is #2 correct? If it is, wouldn't an immediate turn inbound also be correct? What is the definition of "procedure turn"? Does it require flying over a fix to "start"? Or can you be anywhere in the protected area?
The easy solution is to ask for a straight-in clearance, but assume you're not in radio contact or your request is denied.

I'm not sure why you would intercept the localizer outbound, you are already established outbound in holding on the protected side of the course.  Presuming that your right turn at BUHRS offset you sufficiently West to make another right turn inbound, I would commence an immediate descent to at or above 2600, start my right turn inbound to remain within 10NM, then once established on the localizer inbound, descend to at or above 2100 until you intercept the glideslope at BUHRS.  Then fly the rest of the approach as published.


text:
I have a European SPL EASA license - can I fly a Cessna 172 in the US?
I have a European EASA SPL licence with 200 flight hours logged and would like to fly in the US. I understand that a few hours flight training with a US instructor due to differences of rules makes perfectly sense. I also understand that I need a TSA approval.
My question is, will I be able to get a Foreign Private Pilot License issued and fly a Cessna 172?

Yes. But there some hoops to jump through. Check out this post. https://aviation.stackexchange.com/questions/183/i-have-a-european-easa-license-what-do-i-need-to-do-to-be-allowed-to-fly-in-th/227#227


text:
Which 737 generations doesn't have Alpha Vane (Angle of Attack sensor)?
I was wondering if all generations have this angle of attack sensor, especially after recent 737 MAX Lion Air accident.

Are you sure you are asking about the sensor, as opposed to indication of AoA? Practically all airliners have the former, but not all (esp. western ones) have the latter.

Alpha vanes are absolutely essential sensors providing the stall warning. What contributed to the Lion Air flight 610 accident is not the presence of alpha vanes alone, nor the fact one was installed incorrectly. If they fed just stall warning, it would be relatively easy to recognize it as spurious. The main factors were that the Maneuvering Characteristics Augmentation System (MCAS)—kind of unusually implemented stick pusher—does not check for mismatch between the sensors, and that maintenance instructions don't include the alpha vanes as possible cause for the issues from preceding flight.

This is like asking which generation of Civic doesn't have a tachometer. This is not some data you need to feed the computer, but the pilot as well.

@JanHudec The AoA indicator is available on all modern Boeings (including the 737 NG and 737 MAX) as an airline option. If installed, AoA will be displayed in the upper right corner of the PFD.

I have experience on the 737 -100,-200, Classic, Next Gen and the Max8 and even the 707 every one has an AOA sensor.  Different part numbers to be sure but they still have them.
Obviously it senses angle of attack and feeds numerous systems: stall warning and auto slat come to mind immediately.

And to feed the pilot, obviously.


text:
Why do you need to pull HDG knob a320 for selected mode?
If you are flying in nav mode, then atc tells you turn to a different heading, as soon as you change the heading knob the aircraft starts turning, then why do i need to pull it?

What does the FMA say when your rotate the heading knob without pulling?

The FMA remains in NAV mode until you pull the knob out.

When you are in NAV mode the heading indication on the FCU is normally dashed (---*). Then, if you just rotate the knob it first synchronizes the heading to the current heading, then displays the selected heading value and you can pre-select it in the FCU. But it remains in NAV mode and there still is the managed dot after the selected heading value in the FCU (e.g. 270*) until you pull the knob out. When you pull the knob it goes to HDG or TRK mode and the managed dot in the FCU disappears (e.g. just 270 without dot) and only then does the aircraft initiate the turn towards the selected heading.

If you do not pull the knob then the heading value will disappear after a while (45s or so) and the FCU heading will be dashed again.

Once you are in HDG or TRK mode you can just rotate the selected heading knob and the aircraft will immediately turn. You don't need to pull it again.

It's the same for the selected airspeed and selected vertical speed. You can rotate the knob to pre-select a value but you have to confirm the change to selected mode by pulling the knob.

But , why can't I just rotate knob in managed mode and then fly my desired heading without pulling? what happens if i continue to follow atc radar vectors but don't pull it but keep on changing the knob ?

it starts turning as soon as i change the knob, even without pulling it

@SachinChaudhary Are you talking about a simulator or the real aircraft? I'm fairly certain it won't turn in NAV mode when rotating the knob.

FCOM is available online if you want to read the real documentation: https://www.smartcockpit.com/aircraft-ressources/A320-Autoflight.html


text:
Since modern ADF receivers don't rotate, how do they work?
Can you please explain the working of modern electric ADF recievers

Have you seen this excellent answer on how ADF antennas work?

@Bianfable i've seen this. this does not explain the working of the electric adf receiver as well. i want to know how it processes the signals

@mins what i'm really trying to ask is that when the ADF receives the signal, how would the ADF components process this signal electrically to find NULL  - Since there is NO antennae that is able to Rotate physically in space?

At some point a question crosses the threshold from aviation to electrical engineering theory.  If the first linked question doesn’t explain the working of an ADF well enough then we are knocking on that door…. (What are you looking for, a functional breakdown of components on the circuit board?!)

A modern ADF antenna is made either of 4 individual antennas or off two crossed loops.

ADF array, source
The signal strength is the same on all elements, however the distance between elements introduces a delay, which translates into a phase difference on antennas.
The principle is based on comparing the phase on two diametrically opposite elements (or in a loop).
With 2 pairs the system is able to determine the U and V amplitudes of a vector. By combining U and V it is possible to determine the direction of arrival. The accuracy is a few degrees, depending on the size of the array.

Sensing the phase difference (the dark green sine is the phase of the top-left element
This principle is not new, but is now very simple due to modern comparison techniques. It is known as Watson-Watt direction finder:

Original Watson-Watt direction finder using a Lissajous pattern, source


text:
Can a great circle be drawn between any two points on Earth?
If the two points are on the same parallel of latitude, then I can't imagine a great circle through them. Is it right?

https://math.stackexchange.com/questions/2233245/find-a-formula-for-the-great-circle-distance-between-two-points-on-the-same-lati

Related: Explaining a "great circle route" to young civil air patrol cadets

Since latitude and longitude can be arbitrarily assigned, your claim would imply that no great circles could be made.

See also: https://math.stackexchange.com/questions/2356212/why-is-there-always-a-great-circle-through-two-points-on-the-sphere

@StianYttervik Now now. The Earth Science SE would wholeheartedly disagree about the latitudes: After all, they are derived from the poles which are not at all arbitrary. And while Earth Science SE would agree that the International Reference Meridian is entirely arbitrary, History SE would wholeheartedly disagree ;-).

@Peter-ReinstateMonica Well, my claim wasn't that they were arbitrarily assigned, I claimed they can be arbitrarily assigned. A point which rests on the symmetries of a sphere. Since great circles, meridians and latitudes are all presuming the earth-is-a-sphere similitude - I hope such a simple argument will be permitted in evidence.

I believe you are having trouble imagining it because you are allowing the smaller circles formed by the lines of latitude themselves to constrain your thinking.  Consider two points at roughly 45 degrees North latitude - One point is at 0 degrees longitude, and the other is at 180 degrees longitude.
Now how would you connect them?
You could connect them by following the line along 45 degrees North, but that would not follow the great circle.  The great circle route would take you up and over the North pole.
If it helps, take your mental picture of the common latitude “slice”, use the two points as a fixed hinge, then in your mind imagine stretching this ring, and rotating it downwards until the imaginary plane slices through a point at the very center of the earth.  These three points, (the center of the earth plus your two on the surface) define a plane that forms the largest (great) circle that can be made which intersects your points.
If your imagination is insufficient, get a bag of oranges, a pen, and a knife.  Make two dots any where on the orange and then slice through them in a way that creates two equal sized halves.  Keep going until you have freed your thinking.
ADDENDUM:
Federico’s excellent answer provides food for though on the non-spherical nature of our planet. I was curious, so I ran some numbers for a comparison to put things in perspective:
If you circumnavigated the globe around the equator vs a polar route at 420 MPH, after almost two and a half full days continuously airborne the equatorial route would be a mere 6 minutes longer.  That's a 0.1 difference in your logbook after almost 60 hours of flight time.
That's not much.  Probably less than the average time spent taking vectors for landing opposite direction vs a straight-in.  Just an FYI for anyone curious about the practical effect of the equatorial bulge from a pilot's point of view.

So when the FMS calculates the route, it is not planning GC calculations per se, right? It is using some other techniques? It considers the irregularity of the shape of the earth?

@SachinChaudhary, I have no idea how FMS software is written, but unless someone can calculate the difference between GC and Geodesic over the course of a "normal" flight, I'm not convinced it matters. In the C-130 we used to fly polar routes with my navigator friends performing celestial navigation. They were the first ones to introduce me to the concept of GC, and none of them ever mentioned accounting for a less than perfect sphere. But I'm wide open to learning something new if anyone can quantify the net effect in greater detail than my example.

note that circumnavigating the equator does not really mean your error will be representative. two almost antipodal points might even have 0 error. See image on page 5 http://clynchg3c.com/Technote/general/navpaths.pdf

A great circle drawn along a sphere will, with the exception of the one drawn exactly along the equator, intersect all parallels of latitude it passes by exactly twice.
To make it intersect any two points on a specific parallel of latitude is simply a question of alignment of the great circle:

Image source: ResearchGate - DESIGN OF AN EFFICIENT & WEATHER RESISTANT FLIGHT PATH ...
The earth, however, is not a perfect sphere so this answer does not perfectly apply to the case of the earth. See Federico's answer for more detail

That should be longitude in the picture, shouldn't it?

@Auberron The picture shows a great circle, and two lines of latitude: one where the latitude is 0° (the equator), and one where the latitude is 40°N. It doesn't show any lines of longitude.

"A great circle drawn along a sphere will, with the exception of the one drawn exactly along the equator, intersect all parallels of latitude it passes by exactly twice" Except for the points of maximum and mimimum latitude, of course.

Well, they are points, so...

The shortest path between any two point on a sphere follows a great circle. There is exactly one great circle for any pair of two non-identical points on the surface of a sphere, unless they are antipodal, where you have infinitely many great circles (think of the meridians, they go through both poles, and they are all the same). The notion of great circles is most useful when talking about spheres.
Earth has an irregular shape. A spherical approximation is useful and sufficient in very many cases. A better approximation that is indeed used and required for accurate calculations in aviation (like GPS processing) is an (oblate) ellipsoid shape. The shortest path between two arbitrary points on an ellipsoid does not follow a great circle.

Geodesics are the shortest path on a curved surface (e.g. sphere or ellipsoid) - like a straight line on a flat plane. A great circle is the shortest path on a sphere, but not on an ellipsoid (as long as both axes are not of equal length). So every great circle is a geodesic, but not every geodesic must be a great circle (on an ellipsoid for example).

This means that the shortest path between two points on the surface of the Earth is neither a piece of a circle nor even a piece of a closed loop, with the exception of

two points that lay on the Equator (leading to an actual circle)
one of the two points is one of the poles (leading to an ellipse)
two points that differ in Longitude by exactly $180^\circ$ (again leading to an ellipse)

For all other cases, no matter their latitude, the shortest line connecting them will not be a closed line returning to its starting point after only one loop.

Image from Wikimedia (Wikipedia article)
In summary:

Can great circle be drawn between any two points on earth?

Technically speaking, no. Not with a strict definition of "Great Circle", and not with an accurate model of the Earth.

Comments are not for extended discussion; this conversation has been moved to chat.

Consider geometric planes
Yes, and the simplest evidence for that reduces to the notion that you can draw a unique geometric plane between any 3 points not on the same line, so there exists a specific plane for any 2 points on surface of a sphere and its center.
By definition, the intersection of that plane and the sphere is a great circle, and it passes through these 2 points.

Excellent answer - the clearest explanation of why the answer to the question as asked (pedantic "not a perfect sphere" considerations aside) is undoubtedly "yes".

The question appears to be mathematical but also navigational implying a practical side to it.
Lines of latitude on a sphere are technically called "small circles" and it must be appreciated that two points on a sphere may lie on an infinite number of smaller but on only one great circle simultaneously (except antipodal points, of course).  That's the mathematical answer.
Since your post is tagged "navigation", the shape of the Earth is not in actuality a sphere; there are several 3D models of the Earth's shape in use (the World Geodetic System defines them) and these shapes preclude a "great circle" as such.  You can see the WGS quoted on ECDIS* charts used on ships.  This shape is a parameter in the calculations used to plot a course to a destination.  In such a system, a "great circle" must be redefined.  On a  practical note, however, if a ship should arrive with a nautical mile or so of the entrance to a harbour, the crew "done good" so there are limits to how much accuracy you need in calculation and Great Circle Navigation (where the Earth is treated as a sphere) is adequate.  Get that: adequate!  An autonomous ship would need tighter precision.  A crewed ship would not. The crew spies the entrance and manually sails it into the harbour with a pilot on board.  On aircraft, these types of electronic charts save navigators the sometimes mind numbing calculations required for great circle navigation (navigation where the Earth is assumed to be a sphere) which is often okay enough. In range of the airport an aircraft picks up the local radio navigation aids to land.  The trouble with such systems is that crews should still KNOW how to calculate these routes; the equipment tends to take the effort out of it, which is fine.  Until the electronics give out at which time the pilot/sea captain is on his own!  Thus celestial navigation is STILL mandatory for Master's Licenses for sea.  And crews should still know how to calculate them. Yachtsmen still practice them but GPS has been taking all the skilled effort of calculation out of even this culture.
Keep looking at these things, defining words and asking questions and you will find yourself knowledgeable beyond belief in the subject.  And you'll never get lost and if by chance you do, you'll find out where you are soon enough.  Good luck in your studies.
*Electronic Chart Data Information System.

Some proofs:
Imagine taking a rope connecting the two points, and pulling it taut. If the Earth were a perfect sphere and there were nothing for the rope to catch on, the rope would eventually take the shape of a chord of a great circle.
I assume that you agree that given any point other than the North pole, there is a great circle route from the North pole to that point. This gives two more proofs that, given two points A and B, there is a great circle route between them:
Rotate the globe so that A is where the North pole was. By spherical symmetry, there is a great circle route between where A now is and where B now is. We can rotate the globe back to where it was before, and this route will rotate along with it.
Take a great circle route between the North pole and A. Now rotate that circle about the point A. As you rotate the circle, it will cover the whole globe, including B. That is, as you rotate the circle, it will hit B, and when it does, you have a great circle route between A and B.

I think you are getting confused using the lines of lat and long and cross referencing them against great circle routes. On a Sphere it does not matter where you are if you mark 2 spots on a Ball  X and Y .There will be two arcing paths to each of these points both equidistant. In other words. The shortest distance between two points on a plane is a straight line on a sphere its two arcing lines that are the same length and equidistant to each other (Mirror Image).

It sounds like you are saying there are two mirror-image arcing paths that each represent the shortest distance between two points on a sphere. If so, you are mistaken. There's only one great-circle route between two points (unless those points are antipodes), in which case there's an infinite number. Yes, on a flat map that one great-circle route will be an arc, not a straight line, but that doesn't mean that there's a second arc you could drawn on the map, the mirror-image of the first, that would also be a great circle route.  If I'm misunderstanding your answer, please clarify-

And on the globe itself, that one great-circle route appears linear, not arcing, when viewed from directly above a point on the route.  It sounds like you are confusing the curvature you see when looking at a great-circle route on a flap map, with something you'd see when looking at the same route from directly above on the globe itself.

And welcome to ASE, new user-- sorry to be so harsh.  There's no penalty for deleting an answer and submitting a new one to this or some other question!

"flap map" should read "flat map", of course --


text:
Why you have to maintain "S" speed in a320 circuits after a climb?
I've been told that in circuits, after reaching thrust acceleration altitude , you should maintain S speed, after a flaps 1 take off? why is it so? why that particular speed. please explain

Are you flying the whole circuit in config 1?

Yes cfg 1 @Bianfable

S-speed gives good maneuvering margins without being too fast which would make the turns too wide for a close pattern. S-speed automatically adjusts for current aircraft weight, so it decreases the work load to just use that.

When flying in clean configuration, the Green Dot (GD) speed is the recommended speed for approach and holding1 because

it provides sufficient maneuvering margin:

Deviating below GD involves an increase in the drag on the aircraft and would eventually undermine the aircraft’s ability to continue a climb. [...]
Therefore in the clean configuration, the crew should not fly below GD in order to avoid degrading climb performance.

(Airbus Safety First - Control your speed… during climb)

it minimizes fuel consumption:

GD speed is the managed speed target in CONF CLEAN when the FMS approach phase is activated. It is also the recommended speed to extend flaps to CONF 1 and for a holding in clean configuration. [...]
In some phases of flight, GD is computed to minimize drag and thus, the fuel consumption (for example during the HOLD phase).

(Airbus Safety First - Control your Speed… During Descent, Approach and Landing)

However, you said you are flying the circuits in CONF 1 (slats extended). GD speed is replaced by S Speed in CONF 1 (and 1+F). Similarly, F Speed is used for CONF 2 and 3.

In approach phase, S speed is the managed speed target, when in CONF 1 or 1+F. It is the recommended speed to select CONF 2.

(Airbus Safety First - Control your Speed… During Descent, Approach and Landing)
The FCTM also recommends flying holding patterns in CONF 1 at S Speed:

HOLDING SPEED AND CONFIGURATION
If a hold is to be flown, provided NAV mode is engaged and the speed is managed, an
automatic speed reduction will occur to achieve the Maximum Endurance speed when
entering the holding pattern. The Maximum Endurance speed is approximately equal to
Green Dot and provides the lowest hourly fuel consumption.
If the Maximum Endurance speed is greater than the ICAO or state maximum holding
speed, the crew should select flap 1 below 20 000 ft and fly S speed.

(Airbus A320 FCTM - Normal Operations - Holding, emphasis mine)

1 Circuits aren't mentioned in the manuals I have, but they are operationally similar to holding and approach phase.


text:
Can you fly with INS/ IRS at higher lattitudes?
IRS can't be aligned at higher lattitudes ( above 70 deg) , but if once aligned near equator, can we fly till higher lattitudes with IRS in that flight?

Here is some useful information from Boeing regarding navigation and use of IRS/INS in polar regions: https://www.boeing.com/commercial/aeromagazine/aero_16/polar_nav_by_model.html

Yes this would be possible. The limitation is for gyro-compassing only (alignment).
However being technically possible is only one aspect, there are associated regulations. Going into polar regions is not allowed without authorization for commercial aviation. For the US the requirements are described in FAR 135.98 and AC 135-42.
Navigating the polar regions
Aircraft IRS sense true heading, but also provide the magnetic heading. As magnetic heading cannot be determined accurately in polar regions, magnetic heading use is prohibited in polar regions.

Navigating in polar regions prior to FMS wide use required the crews to use a specific navigation grid.

On a FMS equipped aircraft this is transparent, heading is changed automatically from magnetic to true when entering a polar region. Similarly the system uses only one of the IRS instead of 3. More about these details in this Boeing document.

Inertial navigation for polar navigation has been common
This is inertial navigation which allowed the Nautilus submarine to travel under the north pole ice in 1958. This system, a N6A designed for aircraft, was borrowed to USAF. The periodic realignment had to be done using astronavigation and hyperbolic positioning. INS can also be gyro-compassed at high latitude by transfer alignment, like it is done for torpedoes.
INS were used (and are still common) in all missile navigation systems. Most of them had high latitudes routes to either nuke Washington or Leningrad.

So a modern airliner if aligned on equator, would normally be able to fly in polar regions?

On the equator or anywhere south of about 70N or north of about 70S

One detail to add - you not only have to consider normal operations - but also if the INS has some failure and needs to be put into attitude mode - and this realignment might have issues too close to the poles.

@tsg: I covered this a bit in the last paragraph but one, mentioning possible realignment methods. None is available for civil aircraft. INS position could be easily transferred from the GPS receiver, but instantaneous GPS-determined north azimuth is by several orders not precise enough for INS operation.


text:
Question about Eurofighter structure
I was researching for the Eurofighter and I came up with a picture of its structure. I have three questions about its parts. I can't understand the role and the usage of them. Can anyone help me?
These 3 parts are the following:

Radar mounting bulkhead
Strobe light/anti-collision beacon
Engine bleed-air primary heat exchanger

*All of them are circled with a different color from me so to be focused and easy to find them. Also, their definition exists in the picture but I would like to also know, as I mentioned above, their role and their usage.
PICTURE

Thank you for your time in advance!

Radar mounting bulkhead: this is the vertical plate to which the radar antenna is attached. This is just a structure.
Strobe light/anti-collision beacon: this is a bright light that flashes to make the aircraft more visible and prevent collisions.
Engine bleed-air primary heat exchanger: this is a heat exchanger for bleed air from the engine. Bleed air from the engine is hot and needs to be cooled down for various applications, this cooling down is done in the heat exchanger.

It's quite easy these days to find out what an unknown term means. Just go to Wikipedia and enter the unknown term into the search box.